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Language:
Fibonacci
Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn ? 1 + Fn ? 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … An alternative formula for the Fibonacci sequence is
Given an integer n, your goal is to compute the last 4 digits of Fn. Input The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number ?1. Output For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000). Sample Input 0 9 999999999 1000000000 -1 Sample Output 0 34 626 6875 Hint As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
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题意：求非波拉契数列第n项mod10000
思路：数据太大，用到矩阵快速幂。
代码：
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.#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 10000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
typedef vector<ll>vec;
typedef vector<vec>mat;
ll n;
mat mul(mat &A,mat &B)
{
mat C(A.size(),vec(B[0].size()));
for (int i=0;i<A.size();i++)
{
for (int k=0;k<B.size();k++)
{
for (int j=0;j<B[0].size();j++)
C[i][j]=(C[i][j]+A[i][k]*B[k][j])%mod;
}
}
return C;
}
mat pow(mat A,ll n)
{
mat B(A.size(),vec(A.size()));
for (int i=0;i<A.size();i++)
B[i][i]=1;
while (n>0)
{
if (n&1) B=mul(B,A);
A=mul(A,A);
n>>=1;
}
return B;
}
void solve()
{
mat A(2,vec(2));
A[0][0]=1;A[0][1]=1;
A[1][0]=1;A[1][1]=0;
A=pow(A,n);
pf("%lld\n",(A[1][0]%mod));
}
int main()
{
int i,j;
while (scanf("%lld",&n))
{
if (n==-1) break;
solve();
}
return 0;
}
原文地址：http://blog.csdn.net/u014422052/article/details/44684711
